[Akutni]

Are the products of this reaction correct?

[Babcock_Hall]

Would you please provide your thinking first?

[Akutni]

Would you please provide your thinking first?

I think the chlorine is exchanged by the cyanide group. And –OH group reacts with –C≡N group forming the aforesaid product.

[Babcock_Hall]

That is a very reasonable idea.  IMO the only thing left to check is the stereochemical outcome.  I suggest assigning the configurations.

[rolnor]

This does not look like a stable product, if the reaction is done in water, the ester is more likely formed.

[Vidya]

Would you please provide your thinking first?

I think the chlorine is exchanged by the cyanide group. And –OH group reacts with –C≡N group forming the aforesaid product.

Your idea is right.
Chlorine is the better leaving group and OH is very poor leaving group so CN- which is very strong nucleophile  reacts by SN2 mechanism and finally carbon of the CN group is attacked by lone pairs of oxygen and during this mechanism protonation and deprotonation takes place to give that cyclic ring.

[rolnor]

Sorry, the cyclic ester, a lactone is formed.

[Babcock_Hall]

Did the question specify the solvent?

[rolnor]

It could also form the open-chain hydroxy carboxamide, that is more likely if the reraction solution is very concentrated and a closed flask is used.

[rolnor]

It could be a trick question, cyaniode is a poor Sn2 nucleophile and you might get elimination... Or you get the open chain hyedroxy nitrile. Chloride is a very poor leaving group so this reaction is very slow.

[Akutni]

Did the question specify the solvent?

No, but it should be Sn2 reaction according to the instructions.

[Babcock_Hall]

Rolnor made a good point that in water an imine might hydrolyze.  I take a slightly different view regarding the nucleophilicity of cyanide ion.  Its value of n_MeOH against methyl iodide is 6.7, which is comparable to hydroxyl amine.

[rolnor]

I think it depends om solvent, I have not worked with cyanide but I have colleagues who has and they are not to positive about it. But I can be wrong. Chloride is a poor leaving group anyway, the reaction is probably not practical. A secondary chloride is poorer Sn2 electrophile still.

[rolnor]

There is really no reason why the cyanide would rect further than forming a nitrile, nitriles do not react with alcohols under these conditions.