Q1.V.What is index of hydrogen deficiency of C₈H₈?
(a) 1
(b) 2
(c) 3
(d) 4
Q1.VI.Which of the following statements about the mass spectrum of CH₃Cl is correct?
(a) There is one peak for the molecular ion with an m/e value of 45.
(b) There is one peak for the molecular ion with an m/e value of 52.
(c) The last two peaks have abundances in the ratio 1:1 and occur at m/e values of 50 and 52.
(d) The last two peaks are of equal size and occur at m/e values of 50 and 52.
Q2.a.i.Bromobutane, CH₃CH₂CH₂CH₂Br, can be reacted with hot aqueous sodium hydroxide to prepare butan-1-ol: CH₃CH₂CH₂CH₂Br + OH⁻ → CH₃CH₂CH₂CH₂OH + Br⁻. Butan-1-ol produced can be analysed by mass spectrometry. Predict the formula of the fragment ion which is produced at m/e value of 17.
The fragment ion at m/e = 17 is OH⁺ (hydroxyl cation), formed by cleavage of the O–H containing part of the alcohol: mass = 16 (O) + 1 (H) = 17.
Q2.a.ii.Predict two other fragment ions that you would expect to see in the mass spectrum of butan-1-ol and state the m/e value of each ion.
Butan-1-ol, CH₃CH₂CH₂CH₂OH (M = 74), can fragment by: (i) loss of •OH to give the butyl cation, C₄H₉⁺, at m/e = 57; and (ii) C–C bond cleavage next to the OH group (α-cleavage) to give CH₂=OH⁺, at m/e = 31 (loss of C₃H₇• as a propyl radical).
Q2.b.Identify the functional groups responsible for the peaks labelled X and Y. [Same IR spectrum as in Q1.I: Transmittance (%) vs Wavenumber (cm⁻¹), with a broad deep trough labelled X at roughly 3300–2500 cm⁻¹ and a sharp deep trough labelled Y at roughly 1700–1750 cm⁻¹.]
Peak X (broad, strong absorption around 2500–3300 cm⁻¹) is due to the O–H bond of a carboxylic acid. Peak Y (strong, sharp absorption around 1700–1750 cm⁻¹) is due to the C=O (carbonyl) bond of the carboxyl group.
Q2.c.Explain how the two isomers shown in the diagram can be distinguished using IR spectroscopy. [Diagram shows two five-membered ring isomers: compound A is a cyclopentane ring bearing an exocyclic C=O group (cyclopentanone); compound B is a cyclopentene ring bearing an OH group (a cyclic allylic alcohol, cyclopentenol) — both have molecular formula C₅H₈O.]
Compound A (the ketone) shows a strong, sharp C=O (carbonyl) absorption at about 1670–1740 cm⁻¹ and no absorption in the 3200–3600 cm⁻¹ O–H region. Compound B (the cyclic alcohol) shows a broad O–H absorption at about 3200–3600 cm⁻¹ and no carbonyl absorption, with only a weaker C=C absorption around 1500–1680 cm⁻¹ from the ring double bond. The presence/absence of the sharp C=O band versus the broad O–H band distinguishes the two isomers.
Q2.d.i.The molecular formulas of but-1-ene and but-1,3-diene are given below: but-1-ene: C₄H₈, but-1,3-diene: C₄H₆. Find out the IHD for both of these compounds.
Q2.d.ii.Compare the UV absorption spectrum of but-1-ene to that of but-1,3-diene by stating the difference in their λmax qualitatively.
But-1,3-diene has two conjugated C=C double bonds, which brings its π and π* energy levels closer together, so less energy (longer wavelength) is needed for the π→π* transition; it absorbs at a longer λmax (~217 nm, as given for buta-1,3-diene in the text). But-1-ene has only one isolated double bond (not conjugated), so its π→π* transition requires more energy and occurs at a much shorter λmax. Thus λmax(but-1,3-diene) > λmax(but-1-ene), illustrating that greater conjugation shifts λmax to longer wavelength.
Q3.b.i.How will you differentiate between the following using UV visible spectroscopy: [Structure i) a cyclohexene ring (six-membered ring containing one ring C=C double bond) with a conjugated buta-1,3-dien-1-yl side chain (–CH=CH–CH=CH₂) attached directly at the ring double bond, so the ring alkene and the side-chain diene form one continuous conjugated system of three C=C double bonds.]
This compound has three conjugated C=C double bonds (the ring double bond plus the two in the side chain, all in conjugation). Extended conjugation lowers the π–π* energy gap further than in a simple diene, so this compound shows a UV absorption at a longer λmax than an isolated diene (λmax increases by roughly 30–40 nm for each additional conjugated double bond), and its absorption band is typically more intense.
Q3.b.ii.How will you differentiate between the following using UV visible spectroscopy: [Structure ii) a cyclohexane ring (fully saturated, no ring double bond) with a buta-1,3-dien-1-yl side chain (–CH=CH–CH=CH₂) attached to it; only the two double bonds in the side chain are conjugated with each other, and the saturated ring takes no part in the conjugation.]
Because the ring here is saturated (cyclohexane), only the two double bonds of the side-chain diene are conjugated, the same conjugated system as plain buta-1,3-diene. This compound therefore absorbs at a shorter λmax than structure (i) (closer to the ~217 nm value for a simple conjugated diene), since its conjugation is not extended by a ring double bond. Recording the UV-Vis spectra of both and comparing λmax (and absorbance intensity) distinguishes the two: the cyclohexenyl-diene (i) shows the longer λmax, the cyclohexyl-diene (ii) the shorter one.
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