# Why do we need sulfuric acid in creating alkyl bromide and not in creating alkyl iodide from alcohol and hydrogen halide?

i) $\ce{R-CH_2-OH + HBr ->[H_2SO_4] R-CH_2-Br + H_2O}$

ii) $\ce{R-CH_2-OH + HI -> R-CH_2-I + H_2O}$

What role does sulfuric acid play in the first reaction? Why are we not using it in the second one?

The synthesis of alkyl halides from the corresponding aliphatic alcohols using concentrated hydrohalogen acids was investigated by Klein, Zhang and Jiang.[1] They note:

[W]e found that the reflux of 1-butanol ($\pu{2.34 g}, \pu{31.5 mmol}$) with $48~\%$ hydrogen bromide ($\pu{7 mL}$) for $\pu{4 h}$ on a $\pu{120 °C}$ oil bath only gave low yield of 1-bromobutane ($54~\%$) and moderate purity ($93~\%$). The yield of 1-bromobutane was improved to $82~\%$ with $90~\%$ purity by adding additional sulfuric acid ($98~\%, \pu{1 mL}$). Interestingly, we found that 1-iodobutane could be directly synthesized using 1-butanol and hydriodic acid ($\ce{HI}, 57~\%$) by reflux without adding an additional acid in $80~\%$ yield with $98~\%$ purity.

They do not offer any reasoning why switching from $\ce{HBr}$ to $\ce{HI}$ gives better yields. However, we may use our chemical reasoning to deduce the reason. As has been pointed out and explained multiple times on this site, the acidity of hydrohalogen acids increases from fluorine to iodine: $\ce{HI}$ is a stronger acid than $\ce{HBr}$ is. This means that hydrogen iodide should protonate a greater percentage of alcohol molecules than hydrogen bromide — and it requires the collision of a protonated alcohol and the halide anion for the reaction to proceed ($\mathrm{S_N2}$ mechanism).

Remember that all hydrohalogen acids are gases at room temperature and standard pressure and thus need to be dissolved in water to give the actual acidic solution. While sulphuric acid itself may be less acidic than both pure $\ce{HBr}$ and $\ce{HI}$, adding it to the solution will increase the solution’s overall acidity since it comes with no added water. Hydrogen iodide in itself is acidic enough to promote the reaction as the experimental evidence shows so only in the case of hydrogen bromide is additional acidity needed for the reaction to occur.

Interestingly, the authors also state that alcohols with more than five carbon atoms give worse yields due to their low solubility in $\ce{HI}$. This can be overcome by adding phosphoric acid to the mixture. Again, phosphoric acid may not be strong but it features a low water content. The added acidity may protonate more alcohol molecules allowing the $\mathrm{S_N2}$ reaction to proceed. I point to the other answers as to why using sulphuric acid instead of phosphoric acid is a bad idea in the case of hydrogen iodide.

Thus, to sum up:

• hydrogen iodide is sufficiently acidic to protonate the corresponding alcohol and drive the reaction. More acids are not needed.
• hydrogen bromide is not acidic enough to drive the reaction. Adding sulphuric acid protonates the alcohol partially and allows the reaction to proceed.
• in the case of weakly soluble alcohols and hydrogen iodide, phosphoric acid may be used instead since it does not oxidise iodide (while sulphuric acid does).

[1]: S. M. Klein, C. Zhang, Y. L. Jiang, Tetrahedron Lett. 2008, 49, 2638–2641. DOI: 10.1016/j.tetlet.2008.02.106.

I suspect $\ce{HBr}$ is produced in situ from $\ce{KBr}$ and $\ce{H2SO4}$.

Same process isn't particularly suitable for $\ce{HI}$ generation as it reacts with $\ce{H2SO4}$:

$$\ce{2HI + H2SO4 -> H2SO3 + H2O + I2}$$

I actually seriously doubt second reaction will take place at all. Hydroiodic acid can also act as a reducing agent, converting the resulting alkyl iodides to alkanes. Instead of $\ce{HI}$ is often being used $\ce{I2}$ with red phosphorous, or halide salts (Finkelstein reaction).

As @orthocresol noticed in the comment, the following two reactions are more preferable for alkyl iodides synthesis:

$$\ce{R - OH ->[(C6H5O)3P+ CH3I-] R - I}$$

$$\ce{R - OH ->[ (CH3 - S - CH = N+ (CH3)2) I- , THF] R - I}$$

Also, chapter 2.01.5.4 Alkyl Iodides from Alcohols, p. 16 in Comprehensive Organic Functional Group Transformations II (1) accumulates numerous ways of iodoalkanes synthesis:

The adduct of triphenylphosphine with elemental iodine ($\ce{Ph3PI2}$) is a classic reagent used to convert alcohols cleanly and with inversion to iodides. Addition of imidazole is well known to promote this conversion with high yields. Furthermore, treatment of alcohols with triphenylphosphine and cyanogen iodide affords the corresponding iodides in good yields. The reagent $\ce{KI/BF3 * Et2O}$ in dioxane is highly selective and effective for the transformation of allylic and benzylic alcohols to iodides.

A mild and effective procedure for directly converting secondary, tertiary, and benzylic alcohols into the corresponding iodides involves treatment with iodine in refluxing petroleum ether. The reaction proceeds with inversion of configuration:

Alkyl iodides can also be prepared in a single step from the corresponding alcohols upon treatment with the standard reagent $\ce{P-I2}$. A less well used reagent, 1,2-bis-(diphenylphoshino)ethane, can be used to prepare iodides from primary alcohols in the presence of iodide.

Cerium(III) chloride, a Lewis acid imparting high regio- and chemoselectivity in various chemical transformations, can be used in combination with sodium iodide in refluxing acetonitrile to replace a hydroxy by an iodo group:

This method cannot be applied to tertiary alcohols. In this case an alkene is derived by dehydration of the alcohol.

Finally, a classical method for the transformation of alcohols to halides is the well-known two- step procedure via a sulfonate ester, commonly tosylate (p-toluenesulfonate) or mesylate (methanesulfonate).

Treatment of the sulfonate esters with $\ce{NaI}$ or $\ce{LiI}$ gives iodides in good yields.

(1) Katritzky, A. R.; Ramsden, C.; Taylor, R. J. K. Carbon with One Heteroatom Attached by a Single Bond; Comprehensive Organic Functional Group Transformations II; Elsevier Science Ltd, 2004; Vol. 2. ISBN: 978-0-08-044253-2

• Or, for organic chemists, $\ce{PPh3}$, $\ce{I2}$, imidazole does the trick. (Of course, that's beside the point.) – orthocresol Jul 3 '17 at 7:37
• I updated my answer again and this time I have some serious doubts. Would you recommend to ask the doubts as questions after reading my updated answer? – Nilay Ghosh Jul 3 '17 at 11:00
• @NilayGhosh Sure, if in doubt asking questions is a valuable option. I don't get though how DEA acts are relevant for OP (his profile says he is from Bangladesh), neither I understand how the last quoted paragraph is contradictory to what I already mentioned. I also said nothing about isolation of sulfurous acid. Also, I don't think chemiday.com is a valid source for chemical information. They primarily focus on selling ads and have poor moderation (at least they used to). Apart from that, your answer is a great addition. – andselisk Jul 3 '17 at 11:14
• I will wait for a day or two and then I will propose a bounty for more research/answer for this question. This question deserves it and it has made my day. – Nilay Ghosh Jul 3 '17 at 11:15
• @orthocresol and OP: Oh, aquaeous HI does the job well done. I did so (synthesising 1-iodobutane from butan-1-ol) in my PhD thesis and it works like a charm. – Jan Jul 9 '17 at 16:17

First of all, this question is very interesting. Though the question looks very simple but it is very complicated if looked upon carefully. I am just going to shed light on some points in addition to @andselisk's and @Jan's excellent answers.

• @andselisk said that $$\ce{HBr}$$ is produced in situ from $$\ce{KBr}$$ and $$\ce{H2SO4}$$. So, why not write in this way:

$$\ce{R−CH2−OH + (KBr + H2SO4) −> R−CH2−Br + H2O}$$

But the $$\ce{HBr}$$ might further react with $$\ce{H2SO4}$$:

$$\ce{2HBr + H2SO4 → Br2 + SO2 + 2H2O}$$

If we consider this reaction, it might get a little bit complicated. The reaction might not produce alkyl bromide instead the alcohol might gets oxidised to aldehyde and carboxylic acid in presence of bromine water($$\ce{Br2 + H2O}$$) which in turn reacts with alcohol to form esters according to this paper:

$$\ce{CH3CH2OH + Br2 -> CH3CHO + 2HBr}$$

$$\ce{CH3CHO + Br2 + H2O -> CH3COOH + 2HBr}$$

$$\ce{C2H5OH + CH3COOH -> CH3COOC2H5 + H2O}$$

The Reaction Products - The reaction products obtained at varying initial molarities of bromine were determined. The bromine was always quantitatively reduced to hydrobromic acid, and no bromination took place.

Instead of forming alkyl bromide, ester is formed albeit in minute amounts. Other side reaction might take place(See here)

• $$\ce{HI}$$ does react with $$\ce{H2SO4}$$ but not like @andselisk wrote. Sulfurous acid is an unstable aqueous solution of sulfur dioxide and is only detected in gaseous phase. It disproportionate in aqueous solution forming $$\ce{H2S}$$.

When trying to concentrate the solution by evaporation to produce waterless sulfurous acid it will decompose (reversing the forming reaction). In cooling down a clathrate $$\ce{SO2·5.75H2O}$$ will crystallise which decomposes again at 7 °C (forming $$\ce{H2S}$$) Thus sulfurous acid $$\ce{H2SO3}$$ has not been isolated.

Hence, the correct way is:

$$\ce{8HI + H2SO4 → 4I2 + H2S + 4H2O}$$

• $$\ce{KI}$$ and $$\ce{H2SO4}$$ however does not produce $$\ce{HI}$$ because $$\ce{HI}$$ reacts with $$\ce{H2SO4}$$. The reaction produce different products(here and here).

$$\ce{8KI + 5H2SO4 →[\Delta] 4K2SO4 + 4I2 + H2S + 4H2O}$$

$$\ce{8KI + 9H2SO4 →[303-323 K] 4I2 + H2S + 4H2O + 8KHSO4}$$

As @andselisk wrote, $$\ce{HI}$$ is made by reacting iodine reacts with phosphorus to create phosphorus triiodide, which then reacts with water to form $$\ce{HI}$$ and phosphorous acid. This is the way of producing $$\ce{HI}$$ for simple organic synthesis because it is currently listed as a Federal DEA List I Chemical for its use to produce methamphetamine in the United States and it is illegal to buy $$\ce{HI}$$.

$$\ce{3I2 + 2P + 6H2O → 2PI3 + 6H2O → 6HI + 2H3PO3}$$

• Finkelstein reaction is used when the reactant is an alkyl halide not alcohol.
• Yes as @andselisk said, theoretically $$\ce{HI}$$ can be used to convert primary alcohols into alkyl halides through $$\ce{S_{N}2}$$ substitution, in which the iodide ion replaces the "activated" hydroxyl group (water):

$$\ce{HI}$$ is preferred over other hydrogen halides because the iodide ion is a much better nucleophile than bromide or chloride, so the reaction can take place at a reasonable rate without much heating but the problem is that it is an reducing agent and thus would reduce alkyl iodide to alkane. The yield is catalytically increased by using red phosphorus to reduce the formed iodine.

Although harsh by modern standards, $$\ce{HI}$$ was commonly employed as a reducing agent early on in the history of organic chemistry. Chemists in the 19th century attempted to prepare cyclohexane by $$\ce{HI}$$ reduction of benzene at high temperatures, but instead isolated the rearranged product, methylcyclopentane . As first reported by Kiliani, hydroiodic acid reduction of sugars and other polyols results in the reductive cleavage of several or even all hydroxy groups, although often with poor yield and/or reproducibility. In the case of benzyl alcohols and alcohols with α-carbonyl groups, reduction by $$\ce{HI}$$ can provide synthetically useful yields of the corresponding hydrocarbon product ($$\ce{ROH + 2HI → RH + H2O + I2}$$). This process can be made catalytic in $$\ce{HI}$$ using red phosphorus to reduce the formed $$\ce{I2}$$.