Question
Hydrogen bromide, HBr, reacts with but-1-ene.
(a) Identify the type of reaction.
(b) Two products are possible.
(i) Explain the mechanism for the formation of the major product, using curly arrows to indicate the movement of electron pairs.
(ii) Explain why the mechanism results in one product being formed in greater quantities than the other.
(c) Draw the structure of a section of a polymer formed from three monomers of but-1-ene.
(d) Deduce the hybridization of the first two carbon atoms in but-1-ene.
(e) Describe the bonding between the first two carbon atoms in but-1-ene in terms of orbitals on these atoms.
(f) One isomer of \(C_4H_9Br\) can exist as stereoisomers. Draw the three-dimensional structures of the two stereoisomers using wedge-dash notation, clearly showing the relationship between them.
Answer/Explanation
Answer:
(a) «electrophilic» addition/\(A_E\)
(b) (i) 
curly arrow going from C=C to H of HBr AND curly arrow showing Br leaving ü
correct representation of carbocation ü
curly arrow going from lone pair/negative charge on Br- to \(C^+\)
\(CH_3CHBr(CH_2CH_3)\)/correct final product
(iii) carbocation with ethyl group is stabilized
OR
secondary carbocation more stable «than primary carbocation»
greater electron releasing/inductive effect of «two» electron releasing alkyl groups
«on secondary carbocation compared to one on primary carbocation»
(c) 
(d) «both» \(sp^2\)
(e) Any three of the following:
mixing of an s AND two p orbitals forms «three» \(sp^2\) /hybrid orbitals
\(\sigma \) bond is overlap «of \(sp^2\) orbitals» along axial/internuclear axis
OR
head-on/end-to-end overlap «of \(sp^2\) orbitals»
one p electron from each C does not hybridize
\(\pi \) bond is overlap «of p-orbitals» above and below internuclear axis
OR
sideways overlap «of parallel p orbitals»
(f) 
correct isomer
mirror image shown clearly
Question
The structural formulae of two esters of formula \(C_3H_6O_2\) are shown.
(a) (i) Deduce the number of signals you would expect to find in the \(^1\)H NMR spectrum of each compound.
(ii) Outline why infrared spectroscopy is not used to differentiate between the two esters.
Answer/Explanation
Answer:
(a) (i) 
(ii) same types of bonds «present in both molecules»
OR
same wavenumbers absorbed