C16 -- Hydroxy compounds

Question 1

describe how the 3 alcohols with molecular formula C5H11OH could be distinguished using acidified potassium dichromate (VI) and other reagents. there must be at least one positive test for each compound.

P: CH3CH2CH2CH2CH2OH

Q:
CH3CH2CH2CH2(OH)CH2

R:
(CH3)2C(OH)CH2CH3

Answer 1

1. Add K2Cr2O7 (aq) and dilute H2SO4 to each compound and heat. for P and Q, K2Cr2O7 (aq) turns green. for R, solution remains orange.
2. warm P and Q with alkaline aqueous iodine. Q will form pale yellow ppt while P will not form a ppt.
3. Add PCL5 to R. white fumes of HCl (g) will be observed (or used sodium metal. iodoform test cannot be used to identify alcohol R as there is no CH3CH(OH)-.)

Question 2

Explain why phenol forms 2- or 4-nitrophenol with dilute nitric acid but methylbenzene requires concentrated nitric acid in the presence of concentrated sulphuric acid, to form 2- or 4-nitromethylbenzene.

Answer 2

[equations]
in phenol, lone pair of electrons on O delocalise into benzene ring, increasing electron density of the benzene ring making it more susceptible to electrophilic attack.
hence -OH group of phenol is highly activating and hence does not require the strong electrophile of nitronium ion (NO2+) to bring about nitration.

-CH3 group of methylbenzene is much less activating than -OH group/ is only weakly electron-donating and hence requires stronger electrophile NO2+ which is generated with concentrated HNO3 and concentrated H2SO4 to bring about nitration

both -OH and -CH3 groups are 2,4-directing; substitution takes place in C2 and/or C4 positions.

Question 3

arrange the following phenols in order of increasing pKa values, eya.

A: phenol
B: phenol with -NO2
C: phenol with -OCH3

Answer 3

increasing pKa: B, A, C

B has an electron-withdrawing NO2 group which helps to disperse the negative charge on its phenoxide ion, stabilising the phenoxide and making phenol B the most acidic (lowest pKa).

C has an electron-donating OCH3 group which intensifies the negative charge on its phenoxide ion, destabilising the phenoxide and making phenol C the least acidic.

Question 4

explain why 2-bromopropan-1-ol has a lower pKa than propan-1-ol

Answer 4

The presence of the electron-withdrawing bromine atom disperses the negative charge on the conjugate base of 2-bromopropan-1-ol. Conjugate base of 2-bromopropan-1-ol is more stable than the conjugate base of propan-1-ol, hence 2-bromopropan-1-ol is a stronger acid and has a lower pKa than propan-1-ol.

Question 5

explain why phenol has a lower pKa than propane-1-ol

Answer 5

The p orbital of oxygen atom overlaps with the π-electron cloud of the benzene ring. This allows the lone pair on O/negative charge of phenoxide ion to delocalise into the benzene ring, dispersing the negative charge of the phenoxide ion. The phenoxide ion is more stable than CH3CH2CH2O−, thus phenol is a stronger acid and has a lower pKa than propan-1-ol.
(note: you’re comparing the stability of the ion that the original compound has dissociated into, more stable this ion -> less likely to form back the acid -> higher Ka -> lower pKa -> stronger acid)

Question 6

explain the following observation. treat hexan-3-ol with hot conc. H3PO4 produces 4 organic products, but when cyclohexanol undergoes the same treatment, only one organic product is formed

Answer 6

Type of reaction: Elimination (Dehydration)

For hexan-3-ol, there are two possible constitutional isomer products of elimination. Each
constitutional isomer exhibits cis-trans isomerism, giving rise to four organic products in total.

For cyclohexanol, there is only one organic product of elimination. This product does not exhibit
cis-trans isomerism.

Question 7

state 2 observations that you would expect to see when a small amount of aqueous bromine is added to a sample of vanilloid. bromine can be used to confirm that vanilloid contains a phenol group.

Answer 7

Yellow-orange aqueous bromine is decolourised; white precipitate formed.
Note that misty fumes will not be observed. While HBr is also formed in this reaction, the
amounts formed with a small amount of dilute aqueous bromine would not be sufficient to give a visible observation.

Question 8

Suggest two simple chemical tests, each which may be used to distinguish between
compounds A and B. In each case, give the reagent(s) and condition(s) for the test and the observations for both A and B. Do not use any reagents which have been used above.

Answer 8

Test: neutral FeCl3, rt
Observations: violet colouration for A, no violet colouration for B
Test: PCl5
Observations: dense white fumes for B, no dense white fumes for A
Test: K2Cr2O7, H2SO4, heat
Observations: orange solution turns green for B, orange solution remains for A

Question 9

describe and explain the relative acidities of phenol, ethanol and ethanoic acid. [4]

Answer 9

increasing order of acid strength: ethanol, phenol, ethanoic acid [1]

ethanol is the weakest acid because the electron donating methyl group intensifies the negative charge on O in the ethoxide ion, hence destabilising the ethoxide ion. [1]

phenol is stronger acid than ethanol because the negative charge on O is delocalised into the benzene ring, which stabilises the phenoxide ion. [1]

ethanoic acid is the strongest acid because the negative charge on O is delocalised over the -CO2- group, thereby stabilising the ethanoate ion. [1]

Question 10

the pKa values of ethnic acid and carbonic acid are 4.74 and 6.35 respectively. state and explain what you would observe when ethanoic acid reacts with NaHCO3. [2]

Answer 10

CH3CO2H (stronger acid) + HCO3- (stronger base) -> H2CO3 (weaker acid) + CH3CO2- (weaker c base)

effervescence of CO2 gas [1]

ethanoic acid is a stronger acid so it protonates the conjugate base HCO3- to form H2CO3, which decomposes into CO2 and H2O [1]

Question 11

explain why 2-hydroxybenzoic acid has a lower pK1

Answer 11

due to the close proximity of the -COOH and -OH groups in 2-hydroxybenzoic acid, the anion/ conj base is stabilised by hydrogen bonding

Answer 12

for the first compound, the acyl chloride COCl group has 2 electronegative atoms O and Cl attached to that C atom. for the 2nd compound, its chloroalkane group has only one electronegative Cl atom attached to that C atom. thus the COCl C is more electron deficient, more susceptible to nucleophilic attack compared to CH2Cl C.

in chlorobenzene, the C-Cl bond has partial double bond character. this arises from the delocalisation of the lone pair electrons of Cl over to the benzene ring. this C-Cl bond is stronger and does not undergo hydrolysis under normal conditions.