organic 2

If the reaction of a mixture of NaCN and benzyl bromide were run in twice the volume of the solvent, the rate of reaction would nearly be :
a) same
b) twice as fast
c) 1/2 as fast
d) can't say as the nature of reaction is unknown.

10 Answers

1
rishabh ·

i think it should be 1/4th. since SN2 is bimolecular rate depends on conc. of both.
so if volume is doubled =>conc. is halved => 1/2 * 1/2 = 1/4 * initial rate..

71
Vivek @ Born this Way ·

answer says 1/2

1
rishabh ·

sorry!! i didn't see benzyl bromide i took it as phenyl bromide.
then obviously it'll be 1/2 since it is SN1 reaction.
:)

71
Vivek @ Born this Way ·

I'm sure you won't bid so strongly for a SN1 reaction just because it's a 1 degree benzyl carbocation without knowing about the kind of solvent and Nucleophile?

1
mast ·

i think it will favours sn2 mechanism , since the carbocation formed is 1° and also Br is better leaving group than CN and rest depend on solvent being protic or aprotic ....

71
Vivek @ Born this Way ·

So. I guess the question is set without thinking hard.

1
blueivy ·

No, It will be SN1 only. The carbocation will undergo resonance. Also Benzyl halide is preferred for SN1 than SN2. Given in Jagdamba, if you refer that

71
Vivek @ Born this Way ·

The CORRECT answer is not in the options. it'll be SN2, and 1/4 as fast.

1
cvam ·

due to reso sn1

71
Vivek @ Born this Way ·

See here: Written very clearly

http://www.jce.divched.org/jcedlib/qbank/collection/conceptests/rxnnucle.html

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