A-LevelChemistryOrganic synthesis9701/41 · May/June 2023 · Q410 Marks

State the hybridisation of the carbon atoms and the C–C-H bond angle in benzene, C6H6. Explain how orbital overlap leads to the formation of σ and π bonds in benzene.

(a)State the hybridisation of the carbon atoms and the C–C-H bond angle in benzene, C6H6. Explain how orbital overlap leads to the formation of σ and π bonds in benzene.[3]
(b)(i)Compound Z can be synthesised from benzene in three steps by the route shown in Fig. 4.1. [Figure 4.1] Draw structures for X and Y in Fig. 4.1.[2]
(b)(ii)Give the reagents and conditions for steps 1, 2 and 3. step 1 step 2 step 3[3]
(c)Compound W is an isomer of Z. [Figure 4.2] Give the systematic name of W.[1]
(d)Complete Table 4.1 to show the number of peaks observed in the carbon-13 NMR spectrum for W and Z. [Table 4.1][1]

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Mark scheme

Marking points from the official mark scheme (May/June 2023). The mark or mark type is shown in brackets.

[3] M1 bond angle = 120° AND carbons are sp²
[3] M2 σ bonds are formed by end-on-end / head on / head to head / linear overlap of orbitals
[3] M3 π bonds are formed by sideways / lateral overlap of p orbitals
[2] (structure of X, Y)
[3] M1 step 1 (CH3)2CHBr and FeBr3 / Al Br3
[3] M2 step 2 conc HNO3 and conc H2SO4
[3] M3 step 3 Sn and conc HCl
[1] 3,4,5-trimethylphenylamine
[1] compound number of peaks observed W 6 Z 6 Both correct for one mark

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