Item 7 AAMC Test 7 SN1 versus SN2

[MedGrl@2022]

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Which of the following alkyl halides is most readily prepared by a reaction between the corresponding alcohol and concentrated hydrochloric acid?

A) isopropyl choride
B) methyl chloride
C) sec-Butyl chloride
D) tert-Butyl chloride

I chose B because I thought it might be an SN2 reaction and thus B would have the least steric hinderance. AAMC says that the answer is D "Tertiary alcohols react much more quickly with HCl than do other types of alcohols." I see how an SN1 reaction is possible, but why is the SN1 reaction correct versus the SN2 reaction? What should I have been looking for in this problem that would tell me that it is an SN1 reaction?

 

[Postictal Raiden]

Which of the following alkyl halides is most readily prepared by a reaction between the corresponding alcohol and concentrated hydrochloric acid?

A) isopropyl choride
B) methyl chloride
C) sec-Butyl chloride
D) tert-Butyl chloride

I chose B because I thought it might be an SN2 reaction and thus B would have the least steric hinderance. AAMC says that the answer is D "Tertiary alcohols react much more quickly with HCl than do other types of alcohols." I see how an SN1 reaction is possible, but why is the SN1 reaction correct versus the SN2 reaction? What should I have been looking for in this problem that would tell me that it is an SN1 reaction?

I remember when I was doing this question I had the same doubts you had. Finally, I reason out that due to the large size of the nucleophile (chlorine), SN1 is preferred over SN2. Perhaps another rationale could stem from knowing that HCl is a strong acid and it dissociates completely in solution. Therefore, the dissociated protons would be grabbed by the alcohol oxygen to form a good leaving group (water).

 

[Postictal Raiden]

Yeah, I just thought of it and remembered that SN1 is favored in protic, polar environments.

 

[sdm33]

:I got B too, great minds think alike ,

My thinking was protic doesn't guarantee Sn1, sometimes Sn2 use protic solvents.
However, a tertiary alcohol, which would give the most stable tertiary carbocation. A strong acid will protonate the OH, to make it a good leaving group, normally we don't think of alcohol leaving group , but a strong acid makes it it leave. Note that we can't make a primary carbocation. so in this case Sn1 can be secondary or tertiary, and a tertiary alcohol, which would give the most stable tertiary carbocation. Not to say that it will always go Sn1, because if the question is asking specifically for Sn2 than the primary wins.

Also they can ask you H2SO4 as our strong acid, the reaction will go E1 for 3,2
and E2 for 1, it is a dehydration elimination reaction.

 

[MedGrl@2022]

:I got B too, great minds think alike ,

My thinking was protic doesn't guarantee Sn1, sometimes Sn2 use protic solvents.
However, a tertiary alcohol, which would give the most stable tertiary carbocation. A strong acid will protonate the OH, to make it a good leaving group, normally we don't think of alcohol leaving group , but a strong acid makes it it leave. Note that we can't make a primary carbocation. so in this case Sn1 can be secondary or tertiary, and a tertiary alcohol, which would give the most stable tertiary carbocation. Not to say that it will always go Sn1, because if the question is asking specifically for Sn2 than the primary wins.

Also they can ask you H2SO4 as our strong acid, the reaction will go E1 for 3,2
and E2 for 1, it is a dehydration elimination reaction.

Okay I guess it makes sense... I was wavering between the two... but your right SN1 prefers stronger protic solvents because they stabilize the carbocation... even though SN2 polar solvents also increase the rate of SN2... aprotic polar solvents are preferred because protic solvents would stabilize the nucleophile in SN2 reactions and slow down the rate of the SN2 reaction...

What did you mean by the H2SO4? Why would it be an E reaction and not an SN?

 

[dudewheresmymd]

In acidic solution, an alcohol is always in equilibrium with its protonated form. Once the OH on the alcohol is protonated it becomes a MUCH better leaving group as previous post mentioned (H2O is a good leaving group). Now that you have a good leaving group, either SN1 or SN2 can occur depending on the structure. B/c the answer choices include methyl, primary, secondary, and tertiary substrates, however, SN1 is favored among tertiary substrate (answer choice D) b/c this question I believe is referring to the Lucas Reagent (Zncl2/Hcl, leads to alkyl chloride) which reacts MUCH MUCH faster on tertiary (<1 minute to react) than with primary and secondary substates (>5 mins to react) due to the great stability of the tertiary carbocation. Unfortunately, you would only know this from having done lab or learning about it in class, unless there is something in the passage that leads to this conclusion (not sure haven't taken this test yet.)

Also, the other intuition would be (if this is not referring to the Lucas reagent is that if I wanted to make a primary or 2ndary alkyl halide from an alcohol, I usually would use PBR3/SOCL2/ P/I2 (via an SN2 mechanism). Usually, hydrohalic acids (HBR, HI, HCl) are reserved for tertiary substrates in which case tertiary > secondary > primary > methyl substrates--> leading to answer choice D as the best possible answer. Tertiary substrates cannot undergo SN2 so you never use PBr3 or SOCL2 with tertiary alcohols.

Also, keep in mind that halide ions are weak bases (conjugates of strong acids) therefore they maintain their nucleophilicity in acidic solution! 😎