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Does the given compound give Lucas test?

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My teacher says due to the presence of a strong electron withdrawing group the carbocation formed is unstable.

But I say if it undergoes $\mathrm{S_N2}$ mechanism then an electron withdrawing group should increase its reactivity. Why then does this compound not give the test?

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  • $\begingroup$ Do you think Lucas reagent (ZnCl2, dil. HCl) will favor SN2 reaction pathway? $\endgroup$
    – Rahul Verma
    Commented Sep 5, 2020 at 3:06
  • $\begingroup$ @RahulVerma Yes I do. Is it correct. Or do they undergo only SN1 reaction $\endgroup$
    – ishwar b b
    Commented Sep 5, 2020 at 4:28
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    $\begingroup$ Thats a cyanohydrin not a standard alcohol $\endgroup$
    – Waylander
    Commented Sep 5, 2020 at 7:13
  • $\begingroup$ @ishwarbb Side note, for future reference: It is very welcome to illustrate the structure of organic molecules with a sketcher. You may both improve the representation further, as well demonstrate you understand the structure of organic molecules, if the three atoms $\ce{N#C-C}$ are depicted aligned in a straight line. Sketchers often offer to «clean the structure» for you with a click on a dedicated button. $\endgroup$
    – Buttonwood
    Commented Sep 7, 2020 at 12:13
  • $\begingroup$ @Buttonwood ok I will keep that in mind $\endgroup$
    – ishwar b b
    Commented Sep 7, 2020 at 12:37

1 Answer 1

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Lucas reagent follows SN1 pathway. $\ce{Zn}$ is a lewis acid that tends to bond with $\ce{-OH}$ and form a partial positive charge leading to the formation of a carbocation.

According to Hammond Postulate, the stability of the transition state should allow successive formation of product or intermediate. The stability of the positive charge induced on the carbon if stabilised in the transition state should therefore stabilise the carbocation. The rate determining step is the formation of the carbocation.

Since tertiary carbocations are the most stable the carbocation is formed quite easily and the product formation happens quickly.

Similarly, the primary and secondary alcohols, need to form a primary and secondary carbocation, which is not stable. Hence the reaction is slow. Zinc forms a weaker bond with $\ce{-OH}$ and induces a partial charge on them, thus not forming the whole carbocation.

So I think that in such cases where the carbocation is not stable, the reaction is slow, and may not have a a very well defined SN1 or SN2 classification i.e. truly borderline reaction may not occur, due to accompanying facets like solvation and ion pairs.

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