AAMC Fl 1 C/P Q6

[SirButtcheek]

The question is below

I understand why D is correct, but now that I look at it, both B and A are both correct and can serve as possible answers. Why is D the best answer

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[CommyO]

What makes you think the C-O bond is stronger in the different alcohols?

What relevance is that to a SN2 substitution?

the secondary alcohol is more stable (less reactive) than the primary alcohol (more reactive) to undergo SN2

Recall that 2-pentanol is a chiral molecule
any nucleophile that wants to attack the electrophilic carbon bonded to the hydroxyl group must participate in a 'backside' attack; as opposed to either in 1-pentanol indicating no stereochemistry involvement. Therefore SN2 is much more favorable for a primary alcohol compared to a secondary alcohol that can undergo both SN1 and SN2

The hinderance comes from the C-H bonds rotating freely in space

 

[SirButtcheek]

What makes you think the C-O bond is stronger in the different alcohols?

From my understanding, 2-pentanol would have the hydroxyl on a carbon that is directly adjacent to 2 methyl groups. These methyl groups would act as electron donating group thus making the bond stronger. Is this the proper counter argument to make:

"Although it makes the bond stronger, SN1 can still occur and thus the substitution rate didn't drop necessarily."

I still don't understand why C is wrong. Can you provide more clarification, please?

 

[LawToMed88]

There aren’t any methyl groups in 2-pentanol. The -OH is simply on C2, as opposed to C1 on the primary alc.


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[SirButtcheek]

This is what I mean by methyl donating groups, they will act as electron donating groups. Perhaps I am overthinking this problem?

 

[LawToMed88]

Oxygen is more electronegative than carbon (so if anything, -OH is a acting as an electron-withdrawing group), thus accounting for the dipole, but that’s getting way too far into the weeds for this problem.


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[lac5678]

Remember how SN2 reactions are carried out. They happen in a conserved step in which the nucleophile attacks the electrophile from the back and simultaneously kicks off the leaving group (contrasted with SN1 where there is an intermediate). SN2 reactions favor leaving groups that are on primary carbons because this allows easier access for this conserved backside nucleophilic attack due to less spatial or steric hinderance. This is why chemicals like tert-butanol would never go through a SN2 reaction --> how would the nucleohpile reach that electrophilic carbon with all those methyl groups around it? And why chemicals like 1-pentanol go through SN2 reactions --> all that wide open space (also think about the propensity for a primary carbon to be carbocation intermediate..never!!)

1-pentanol has a primary OH, 2-pentanol has a secondary OH and thus 1-pentanol would be favored and the reaction would occur faster