16

The rate of Aromatic substitution depends upon the activity of the aromatic system, because when the collision happens the aromatic system has to donate electrons to an electrophile. In the example above you have used nitro-benzene, which is very strongly deactivated due to the $\ce{-NO2}$ group and thus the ortho and para positions are completely blocked. ...


15

During a electrophilic aromatic substitution, it is always possible to have multiple substitutions during one reaction. However, your example is not the ideal one for a discussion. As noted, Oscar Lanzi has questioned even aromatic ring of nitrobenzene is active enough to give even one $\ce{Br}$ substitution. To clear that, I have found a reliable reference: ...


12

Briefly an electronically excited state is produced (distinct from the ground state) and if the uv energy is sufficient it is produced above its dissociation energy, which is less than in the ground state. Alternatively in the excited state there is another excited state that the energy from the first one crosses over into and this dissociates. The figure ...


11

This answer is wrong, the answer should be the carbamate 4. The reaction intermediate is the isocyanate which is captured by MeOH to give the carbamate (std Hofmann mechanism below from masterorganicchemistry.com). For a Hofmann in MeOH see this Org Syn procedure here


10

The $\mathrm{S_N1}$ or $\mathrm{S_N2}$ reaction of alkyl halide with nitrite ion is complicated one since nitrite is an ambident nucleophile. For instance, it was reported that silver nitrite reacts with alkyl halides to give nitrites while sodium nitrite gives more nitroalkane than alkyl nitrite (based on Hard Acid/Hard Base concept; Ref.1): $$\ce{AgNO2 + R-...


10

As you have mentioned, the basic Fischer operations are: Vertical positions are below the plane of paper and horizontal positions are Above the plane of paper, thus you have already remember that when working with Fischer projection. What you have to remember about chiral compounds are: If you switch two groups, you get the epimer and if you switch two other ...


10

This reaction should be leaning towards SN1 because the approach of $\ce{H2O}$ for a back attack to the ABMO of the $\ce{C-Br}$ is heavily hindered since the neighborhood is congested due to the small ring of cyclobutane. Once cation formation occurs the reaction will interestingly, result in a ring contraction because the carbocation formed would be very ...


10

Anhydrous $\ce{AlCl3}$ is used because it reacts violently with water details here and you then no longer have an aluminium compound that will react. $\ce{AlCl3}$-catalysed reactions occur because it is a Lewis acid definition here and readily accepts electron pairs. Water readily donates an electron pair which stops any other molecule from combining with it....


10

Further to my comments above, I am going to offer my opinion that the book is wrong and the epoxide is not opened by sodium hydride as depicted. I have failed to find any literature examples of such a reaction and, in 40+ years as a synthetic chemist, I can recall no reactions in which $\ce{NaH}$, unmodified by other reagents, acts as a source of ...


9

Here is my assessment of the mechanism. You are correct about the Dess-Martin periodinane (DMP) oxidation of 1 to 2. The resulting ketone 2 can also exist as the enols (E)-3 and (Z)-4. Any one of the three can undergo a retro-Diels-Alder-like thermolysis, but I have chosen (Z)-4 for the convenience of cyclizing ketene 5, which has lost acetone in the ...


9

You deprotonate the alcohol, and then flip bonds to create the enolate. There are two enolates possible, so swapping bonds gives you the terminal one, which can attack the aldehyde to form the 5-ring. Elimination of water gives you the product.


9

Unlike other reducing metal hydrides (e.g., $\ce{NaBH4}$ and $\ce{LiAlH4}$), diisobutylaluminum hydride (DIBAL-H) is a liquid at room temperature and dissolve in many hydrocarbons such as toluene and hexanes, which also have very low freezing points. For example, hydrocarbons toluene and hexanes both have freezing points around $\pu{-95 ^\circ C}$. Thus, ...


9

It is complicated, as this paper from Olah et al. here shows, but this passage offers a possible explanation: With the reactive nitronium salts, the isomer distribution of the nitration of anisole shows the highest ortho/para ratio (2.7-2.4), reflecting the "early" (i.e., starting aromatic-like) nature of the transition state of highest energy. ...


9

This is a rather unusual and interesting case of aromaticity, which has been given a special name: homoaromaticity. The Wikipedia page does a quite nice job of explaining what's going on. As you state, protonation of cyclooctatetraene generates a $\ce{C8H9^{+}}$ cation containing an $\mathrm{sp^3}$-hybridised carbon atom between all the other $\mathrm{sp^2}$-...


9

The abnormal Beckmann rearrangement differs from the normal one that there is a nitrile as product of reaction. As compiled by William Reusch (formerly Michigan State University) in his freely accessible virtual textbook organic chemistry, the assumed mechanism of the abnormal Beckmann rearrangement is e.g., (image credit) For comparison, the normal ...


8

I think the easier way of doing this is to epoxidise the diene with 1 eq of a peracid epoxidation with peracids . The standard reagent m-Chloroperoxybenzoic acid (MCPBA) breaks your rule of organic molecules with 3 carbon atoms or fewer, but you can also do it with peracetic or pertrifluoroacetic acid preparation of peracetic acid. You will inevitably get ...


8

Yes, I believe the rearrangement can happen. However, the product from the rearrangement is probably not the major one. The closest reaction to what you have drawn that I could find on SciFinder is this: The paper is mentioned below in the reference. In the paper, the researchers performed similar reactions of alkenes with $\ce{I-Cl}$ and in cases where ...


8

There are two possible eliminations that can happen (see the image below) As you can see, we then have two final products depending upon which $\ce{H}$ is eliminated. From the image 1a is cis and 2a is trans. Now, how do we decide which one is the major product? This can be based on looking at the stability of the transition state. This means we have to ...


8

It depends on the reaction conditions. I suspect the answer the question setter wants to see is phthalic acid, but the true answer is none of these. I have found multiple references (such as this US Patent and this Org. Syn prep) that refer to the production of naphthoquinones by oxidation of naphthalenes with high valent Chromium reagents under acid ...


7

You may have confused barium peroxide with barium metal. The metal can indeed displace hydrogen from hypochlorous acid, not to mention (for a metal as reactive as barium) from the water in which the acid is dissolved; but metal peroxides will produce oxygen and water from the acid. So your missing product is water, not hydrogen. Tricky half-reactions If we ...


7

That 1,2-diodopropane is unstable does not indicate whether it is as the neat halide or in solution. Breaking (BDE) two $\ce{C-I}$ bonds costs $\pu{+114 kcal/mol}$ while forming the $\pi$-bond of propene is worth approximately $\pu{-62 kcal/mol}$ and the formation of iodine, $\pu{-36 kcal/mol}$. The net reaction is endothermic by $\pu{+16 kcal/mol}$. The ...


7

As I noted in a Comment that there should be no significant difference between the use of ethanol or isoamyl alcohol in the reduction of naphthalene with sodium in that both alcohols are primary. The difference lies in the conditions of the reaction. The sequence is a classic case of kinetic vs. thermodynamic conditions. The reaction conditions in the ...


7

I appreciate Waylander's answer, but it did not address the OP's curiocity of why the base abstracts the proton from only ortho-position. Yet, Waylander correctly pointed out that since the aromatic $\pi$-system is at right angle to the triple bond (the second $\pi$-bond of the triple bond), the ability to donate a lone pair to the ring by a substituent (...


7

Reading this question, I realized that OP is very new to organic chemistry, and in need for learning a lot about electrophilic aromatic substitution reactions. Thus, I recommend that OP should concentrate on the electrophilic aromatic substitution reactions and read the chapters of OP's textbook dedicated to that subject. Said that, I'm going to answer ...


7

This confusion arise because OP did not look at the energy profile of $\mathrm{S_N}$1 mechanism. The first order kinetics of $\mathrm{S_N}$1 reactions (e.g., reactions given in the question) suggests a two-step mechanism in which the rate-determining step consists of the ionization of the substrate (here, it is an alkyl halide), as shown in the following ...


7

Zn reacted with the allylic bromide to create an allyl zinc species. This is known as the zinc-mediated Barbier- Grignard reaction wikipedia. The allyl zinc species then reacts nucleophilically with the benzaldehyde example here. Reaction can take place through either end of the allylic system with regioselectivity influenced by the choice of reaction ...


7

It will depend on strict control of the equivalents of reductant, your work-up conditions and your control of temperature during the workup. The intial product of the reduction is the aluminium alkoxide. If you use a proton source such as $\ce{HCl (aq)}$ or $\ce{NH4Cl (aq)}$ to quench the alkoxide and keep the reaction mixture cold then you will get the 4-...


7

In the given compound the $\ce{-OH}$ group and the $\ce{-NH2}$ and later $\ce{N2^+}$ group are not in anti position with respect to each other. Equatorial positions in cyclohexane are not anti with respect to each other. Below I have drawn newmann projection of the reactant which should help you visualise that a carbon (encircled) is instead in anti-relation ...


7

This was going to be a comment but it got too long. TL;DR: - This is not an answer, rather a justification for why the question is (probably) wrong. This paper (linked by @Rishi) gives us experimental evidence that under action of concentrated $\ce{H2SO4}$ hydrogens are exchanged from paraffins in the following fashion: $$ \begin{align} \ce{(CH3)2CHCH3 + ...


7

Similar to to last step of aldol condensation to give $\alpha, \beta$-unsaturated ketone, the last elimination step at sought reaction is also supposed to undergo elimination of water molecule through $\mathrm{E1_{cb}}$ mechanism. To refresh your memory, the definition of the $\mathrm{E1_{cb}}$ mechanism is as follows: An elimination reaction mechanism ...


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