I want to ask a question about the hydrolysis of haloalkanes to form alcohols.

I have been presented with a flowchart diagram of possible reactions of haloalkanes below:

flowchart diagram

Now, I have been presented with the question below:


I have identified the reagent I relevant to the question as NaOH (aq)

First, I need to compare 1-bromobutane and 2-bromo-2-methylpropane.

1-bromobutane 2-bromo-2-methylpropane

I can see that there are 3 alkyl groups in 2-bromo-2-methylpropane and only 1 alkyl group donating electrons to the Carbon atom attached to the Bromine.

So, during the reaction, the carbocation intermediate formed by reaction of NaOH with 2-bromo-2-methylpropane will form be more stable than the carbocation intermediate formed with 1-bromopropane (application of Markownikoff's rule) as alkyl groups are electron-donating groups and the three methyl groups will stabilise the positive charge on the carbon atom to a greater extent in 2-bromo-2-methylpropane.

However, I am slightly confused.

Why does this mean that the rate of reaction with 2-bromo-2-methylpropane is faster than the reaction with 1-bromopropane?

  • 2
    $\begingroup$ Your flowchart should include reagents. And phew, there’s a lot hidden in that question D= $\endgroup$ – Jan Jan 1 '18 at 17:37
  • $\begingroup$ @Jan they asked me in the previous question to identify the reagent but I have mentioned that I identified it just now in the edit above, and yes certainly there is a lot hidden, but it is a good thinking question, something I enjoy looking at! $\endgroup$ – vik1245 Jan 1 '18 at 17:40
  • $\begingroup$ With ‘question’ in my previous comment I meant the text you posted on this site. Meaning an answer will have to address quite a few things. $\endgroup$ – Jan Jan 1 '18 at 17:49
  • $\begingroup$ I can't exactly remember where I read this, but in case of substitution of halogen in a haloalkane by $\ce{OH-}$, the overall rate of reaction is tertary>secondary>primary. Even though tertiary reacts via $\ce{S_N1}$ and primary and secondary mostly via $\ce{S_N2}$. $\endgroup$ – Shoubhik R Maiti Jul 25 '19 at 15:36
  • $\begingroup$ Don't know if this is a general trend in case of substitutions on alkane, and no idea why this happens $\endgroup$ – Shoubhik R Maiti Jul 25 '19 at 15:38

Your Answer

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

Browse other questions tagged or ask your own question.