Amines Question AQA

i’m not sure if the images attached but here they are

Hi,
I don’t understand the first step of this question. I’ve attached the mark scheme as well but I don’t understand how they went from methylbenzene to C6H5CH1Cl. I’m assuming it’s a free radical substitution mechanism as the reagents and conditions is Cl2 with UV light but why does that form C6H5CH1Cl instead of C6H5CH2Cl? Also, why is it more correct to use Cl2 instead of Br2 (is it just because chlorine is more electronegative?) I understand the steps after the first one

So you are starting from methylbenzene (C6H5CH3) and they want you to make 2-phenylethylamine (C6H5CH2CH2NH2).
The plan is to extend the carbon chain by 1 - this should immediately prompt you to think KCN in an ethanol/water mixture will be one of the steps in the synthesis (since this is the only way the AQA syllabus teaches you to extend a carbon chain by 1). However, this prompts two questions.
The easier question is how do we go from a nitrile (the product of the reaction with KCN in ethanol) to an amine - simply put, you need to carry out a reduction. LiAlH4 in dry ether works well for reducing nitriles, so it would be my instinctive answer.
The harder question is how do we introduce a group that can be replaced by the cyanide ions to make the nitrile in the first place - what group, and how do we plonk it on the correct carbon?
If you look at the bit we are extending, you’ll see it’s a methyl group - analogous to an alkane. If you recall your reactions of alkanes (thankfully there aren’t a lot), you have combustion and radical substitution.
Combustion is no good, but radical substitution is perhaps useful - place a halogen on the methyl group and you should be able to substitute it. This means using either Cl2 or Br2 under UV light. You’ll also be pleased to know that the benzene ring should be pretty much impervious to reaction under these conditions.
Although the question itself doesn’t ask this, do recall that radical substitution isn’t a great synthetic method as you get a lot of side products (multiple halogenations, termination of two C6H5CH2• radicals blah blah blah).
So we have:
(1. Cl2 or Br2 in UV light to make C6H5CH2X
(2. KCN in an ethanol/water mixture to make C6H5CH2CN
(3. LiAlH4 in dry ether to make C6H5CH2CH2NH2

The reaction is indeed free radical substitution
The product is C6H5CH2Cl - not C6H5CH1Cl. It’s a typo.
It’s not more correct to use Cl2 instead of Br2 - both are equally acceptable, but chlorine is the more commonly encountered reagent as it tends to be shown as an example when first teaching the topic of radical substitution.