| Qu 1: |
(a) Since 1-propanol would be the anti-Markovnikov alcohol,
use BH3 then H2O2 / NaOH |
|
(b) Need to add a C, so open epoxide with an organometallic reagent,
e.g.
CH3MgBr in Et2O or THF, followed by an acidic work-up. |
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(c) Right number of C, so simple reduction using either LiAlH4
in THF then acidic work-up or NaBH4 in EtOH |
|
(d) Right number of C, so simple reduction using LiAlH4
in THF then acidic work-up. NaBH4 will not reduce RCO2H |
|
(e) Reduction of a propanoate ester with LiAlH4 in THF will
give the primary alcohol. |
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| Qu 2: |
(a) Since 2-propanol would be the Markovnikov alcohol, use aq. H2SO4
/ heat. |
|
(b) Right number of C, so simple reduction using either LiAlH4
in THF then acidic work-up or NaBH4 in EtOH |
|
(c) Need to add a C, use an organometallic reagent, e.g. CH3MgBr
in Et2O or THF, followed by an acidic work-up. |
|
(d) Right number of C, just need to change the functional group. Use
aq. NaOH. |
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| Qu 3: |
(a) cis-1,2-cyclohexanediol |
|
(b) trans-1,2-cyclohexanediol |
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(c) (R,R)- and (S,S)-2,3-butanediol |
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(d) meso-2,3-butanediol |
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(e) (R,R)- and (S,S)-2,3-butanediol. Compare this to reaction (d),
changing the stereochemistry of the overall additon process to anti means
that the products will be diastereomeric with those in (d). |
|
(f) They are diastereomers. Since the addition is syn, the relationship
is defined by the relationship of the alkene starting materials. |
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| Qu 4: |
(a) Thionyl chloride converts ROH to RCl, so benzyl chloride, C6H5CH2Cl. |
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(b) A Williamson ether synthesis, giving C6H5CH2OCH2CH3 |
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(c) Under aqueous, the product will be benzoic acid, C6H5CO2H |
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(d) With the more selective oxidation conditions, the product will
be benzaldehyde, C6H5CHO |
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(e) An ester preparation giving benzyl ethanoate, CH3CO2CH2C6H5 |
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| Qu 5: |
In each case, disconnect the alcoholic portion from the acid portion: |
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| Qu 6: |
Periodic acid, HIO4, causes oxidative cleavage of 1,2-diols. |
| |
In the case of a cyclic diol, this would break the ring. The
product here would be 1,6-hexanedial.
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| Qu 7: |
The best method for ether synthesis is the Williamson method from an
alcohol and an alkyl halide. |
|
 |
(a) This is the best disconnection to a primary halide (good SN2) and
the tertiary alcohol. The tertiary bromide would be prone to elimination. |
| (b) This is the only choice since aryl bromides do not undergo substitution,
we must make it the alcoholic component. |
| (c) A symmetrical cyclic ether so we can disconnect either C-O bond,
it makes no difference. |
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