JEE Chemistry Club

10:22 AM

In this question, A is given to be the correct answer, but won’t the major product be formed when the cation attacks para position(wrt -OCH3 group)?

10:45 AM

@RobinSingh Ring strain would be very high in such a product. try drawing its geometry..

11:02 AM

@Safdar Wouldn’t that be a six-membered ring?

Wait a sec, I’ll draw

Avyansh Katiyar

11:40 AM

O shit what was this book

I've forgotten

rash

11:56 AM

@RobinSingh What reaction is this? Sorry guys really poor in chemistry

Is it friedel-crafts?

12:51 PM

@AvyanshKatiyar Himanshu pandey

@rash Yes

1:48 PM

What would be the name of that compound if we attach another benzene ring at para postion to biphenyl and more importantly what would be its 3d structure

2:29 PM

Hi all! I have most probably a dumb question...

@Priya A modification of haloform..

Here I know the three alpha hydrogens get replaced by chlorine. But how can treatment with aqueous acid give carboxylic acid? I don't think the bond between this group and benzene can be easily broken.

@Safdar Hi. Thanks. I was typing my question.

@Priya 1) Did you get the answer? 2) Aqueous acid is for acidic hydrolysis and base is used for the bond cleavage.

I understood the substitution of chlorines. But this is probably the first time I come to know that even benzene - group bond can be broken in a single step, unlike EAS.

Is there any name for this type of reaction?

@Priya It is just haloform of methyl ketones.. Nothing special.

Use I2 instead and it becomes I2/NaOH followed by acidic hydrolysis.. (Seem familiar?)

2:41 PM

Ok. Iodoform! Thank you :)

I was amazed by the bond cleavage with benzene...

@Priya wdym? isn't it on the other side?

@Safdar Didn't get it. I can just understand that the three hydrogens of methyl ketone will be replaced by chlroines. After that I don't know how the bond cleaves. Will it be similar to Sn1?

If you don't mind can you please give a link showing a mechanism for this reaction?

Haloform reaction

The haloform reaction is a chemical reaction where a haloform (CHX3, where X is a halogen) is produced by the exhaustive halogenation of a methyl ketone (RCOCH3, where R can be either a hydrogen atom, an alkyl or an aryl group), in the presence of a base. The reaction can be used to transform acetyl groups into carboxyl groups or to produce chloroform (CHCl3), bromoform (CHBr3), or iodoform (CHI3) and also cyanide. == Mechanism == In the first step, the halogen disproportionates in the presence of hydroxide to give the halide and hypohalite (example with bromine, but reaction is the same in case...

@Priya It becomes a very good leaving group