5
$\begingroup$

Arrange the following according to reactivity towards nucleophilic addition reaction:

enter image description here

I have already seen this post regarding the reactivity. It helped me to some extent.

$\ce{Cl}$ increases the positive charge on the carbonyl carbon increasing its reactivity.

So (B) > (A) > (the remaining 3)

Further, aldehydes are more reactive than ketones. But I should also consider steric factors. In compounds (C), (D), and (E), compound (D) has highest carbocation stability but also has highest steric hinderance.

Compound (C) has lowest steric hinderance but low carbocation stability as well. How can I arrance (C), (D), and (E)?

To all those who say steric effects dont play a role: This is taken from page 147 of organic chemistry book by Clayden:

The same structural features that favour or disfavour hydrate formation are important in determining the reactivity of carbonyl compounds with other nucleophiles, whether the reactions are reversible or not. Steric hindrance and more alkyl substituents make carbonyl compounds less reactive towards any nucleophile; electron-withdrawing groups and small rings make them more reactive

Improve this question
$\endgroup$
7
  • $\begingroup$ It will be b>a>c>e>d In nucleophillic addition reaction carbocation doesn't form. You Compare only by electrophillic nature and steric hinderance. $\endgroup$
    – Vaibhav
    Dec 25, 2015 at 9:29
  • $\begingroup$ Also remember that formaldehyde is the most reactive aldehyde due to least steric hinderance $\endgroup$
    – Vaibhav
    Dec 25, 2015 at 9:39
  • $\begingroup$ B and A are not acid chlorides. $\endgroup$
    – orthocresol
    Dec 25, 2015 at 11:17
  • $\begingroup$ Sorry. But the Cl increases positive charge on the carbonyl carbon. I will edit the question $\endgroup$
    – Aditya Dev
    Dec 25, 2015 at 11:19
  • $\begingroup$ There's no use of carbocation stability if the bulky groups prevent nucleophiles from attacking there, as per molcalc the order should be $C>E\geq D>B\geq A$ $\endgroup$
    – Sujith Sizon
    Dec 25, 2015 at 13:46

1 Answer 1

Reset to default
-3
$\begingroup$

Carbocation doesn't come into the picture at all, when we are discussing about nucleophilic addition reactions. Steric hindrance will also not play a role because the compounds are planar (attack is going to take place on $sp^2$ hybridised carbon). The nucleophile is going to attack only from the top and bottom* as shown:

enter image description here

The only factors that come into picture in the options are the electron withdrawing or releasing nature of the groups attached to the carbonyl carbon. If the group is $e^-$ withdrawing in nature, the deficiency on the carbon atom increases, making it susceptible to nucleophilic attack.

So, the correct order is : B > A > C > E > D

P.S.:* Since the attack is from top and bottom, both of which are equally likely, the resulting product is a racemic mixture (if the carbon on which attack happens is chiral after the attack).

Improve this answer
$\endgroup$
3

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.