7
$\begingroup$

The reactions of primary and secondary alcohols with halogen acids ($\ce{HX}$, where $\ce{X}$=halogen) require the presence of a catalyst, $\ce{ZnCl_2}$. With tertiary alcohols, the reaction is conducted by simply shaking with concentrated $\ce{HCl}$ at room temperature. Which mean that, tertiary alcohols are more reactive than primary and secondary alcohols. The reason is said to be, due to the inductive effect.

I don't know how the inductive effect would explain that, tertiary alcohol is more reactive than primary and secondary.

Improve this question
$\endgroup$
10
$\begingroup$

Secondary and tertiary alcohols react with strong acid (and bases) such $\ce{HX}$ following SN1 mechanism.

$$\ce{ROH <=> HX + ROH_2^+ + X^-} \tag{1}$$ $$\ce{ROH_2^+ <=> R^+ +H_2O} \tag{2} $$ $$\ce{R+ +X- \rightarrow RX} \tag{3} $$

In these reactions, the rate determining step is the formation of the carbocation (2). So the successive attack of the nucleophilic chlorides is dependent on that.

enter image description here

Here is how the inductive effect comes into play. The inductive effect of $\ce{R}$ groups of the alcohols stabilize the carbocation (figure show the stability of carbocations from the more stable to the less) and then increase the chances of a nucleophilic attack!

Improve this answer
$\endgroup$
7
  • $\begingroup$ @G M.Thank you for the answer. Is more stable carbocation, more reactive? if so. How? $\endgroup$ – Sensebe Nov 10 '13 at 7:08
  • $\begingroup$ @CURIE just think: a specie to react with a reagent must be present, a stable carbocation has more possibility to be present rather a unstable carbocation so more possibilities mean more chances that the reaction take place. You're welcome! $\endgroup$ – G M Nov 10 '13 at 7:36
  • $\begingroup$ @G M.I haven't yet understood, how the presence of more no of carbon atoms in tertiary alcohols, make them more reactive than primary. What I thought was, if tertiary is more reactive than primary alcohol, it means that carbon oxygen bond is weaker in tertiary alcohol than in primary, which is making the -OH bond in tertiary alcohol to break easily than in primary, when a nucleophile is going to attack. My doubt is, how C-O bond is going to be weaker in tertiary than in primary? $\endgroup$ – Sensebe Nov 10 '13 at 8:39
  • $\begingroup$ @CURIE if tertiary is more reactive than primary alcohol, it means that carbon oxygen bond is weaker in tertiary alcohol than in primary What are the bases of this assumption? Reactivity is a complex topic related to thermodynamic and kinetic, in this case the slow step (what determinate the velocity of the reaction) is the formation of carbocation so a more stable carbocation (due the inductive effect) is thermodinamically favorite rather an unstable carbocation. Try to cite the exact phrases of the book so we can see the context more clearly. $\endgroup$ – G M Nov 10 '13 at 9:05
  • $\begingroup$ @G M. I agree with all your explanation that, slow determining step is going to determine the speed of reaction. I assumed that C-O bond to be weaker in tertiary alcohol than in primary or secondary, or in your view inductive effect is making the C-O bond to break easier in tertiary than in primary and secondary. Because, if you think C-O bond to be of same strength in all tertiary, secondary, and primary. When nucleophile approaches carbocation in slow step, C-O bond in tertiary would have same strength (inspite of inductive effect) as in primary or secondary, and reactivity would be same. $\endgroup$ – Sensebe Nov 10 '13 at 11:17
0
$\begingroup$

inductive effect in tertiary carbocation is much more than primary or secondary carbocation since there are 3 R groups in tertinary carbon. so (-)inductie effect by R groups reduces the (+)charge on carbocation. so it will become more stable there for it is thermodynamycally favorite. so the reactivity is heigher than primary or secondary alcohol.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Without stating that the reaction mechanism changes from predominantly $\mathrm{S_N2}$ to predominantly $\mathrm{S_N1}$ when going from primary to tertiary, this is a pretty useless answer. $\endgroup$ – Jan Nov 9 '17 at 12:47
0
$\begingroup$

Tertiary alcohols are more reactive because the increased number of alkyl groups increases $+I$ effect. So, the charge density on carbon atom increases and hence around oxygen atom. This negative charge density tryna push the lone pairs on oxygen atom away. So, the oxygen atom wanted to leave the carbon atom as soon as possible. Hence, the cleavage of $\ce{C-O}$ bond becomes easier. So, a tertiary carbocation is formed which is more stable than secondary and primary. Hence, their formation is favoured by immediate breakage of $\ce{C-O}$ bond in tertiary alcohol. This shows that they are more reactive.

Improve this answer
$\endgroup$
2
  • $\begingroup$ No, this explanation is very wrong. $\endgroup$ – Jan Nov 9 '17 at 12:48
  • $\begingroup$ how jan..........??? $\endgroup$ – prince Nov 9 '17 at 18:27
0
$\begingroup$

This is because when number of electron donating alkyl group on OH-bonded carbon atom increases, polarity of carbon oxygen bond also increases, which further facilitates the cleavage of carbon oxygen bond. Therefore reactivity increases.

Improve this answer
$\endgroup$
-4
$\begingroup$

Primary alcohols react via $S_N2$ mechanism which involves the carbocation formation but the secondary and tertiary react via $S_N1$ which is elementary and no carbocation is formed. hence it appears that both primary and tertiary alcohols should react rapidly but it is not so.

the $\ce{C-O}$ bond weakens as the no of electron donating groups on the alpha-carbon increases this is due to corresponding increase in negative charge developed on $\ce{O}$ atom **

thats why tertiary alcohols are more reactive.

** further $\ce{ZnCl2}$ is a Lewis acid which coordinates with $\ce{O}$ atom and weakens the bond thus it need to be used for primary and secondary alcohols.

Improve this answer
$\endgroup$
1
  • 2
    $\begingroup$ Welcome to Chemistry.se! A SN2 mechanism would not involve the formation of a carbocation, because it is concerted, while in the SN1 reaction a carbocation is formed, as it is stepwise. $\endgroup$ – Martin - マーチン♦ Jan 5 '15 at 3:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.