Tutorial 7:1
∆G� = − 91.3 kJ mol‾� (Temp for standard state is 25�C). ∆H� = − 75.1 kJ mol‾�. So the net energy from the breaking of the C−Cl bond and the formation of the C−O bond is − 75.1 kJ mol‾�. Meaning the C−O is stronger than the C−Cl. This is the driving force for the reaction. Back
Tutorial 8.1
2-hexanol or hexan-2-ol. Chemists would think that you are uncouth if you used 4-hexanol or hexan-4-ol. Back
Tutorial 8.2
CH2=CH2 +
H2SO4
H+
→CH3−CH2−OSO2OH
HO−H +
CH3−CH2−OSO2OH
→
CH3−CH2−OH +
H2SO4
Back
Tutorial 8.3
The oxide of an alkane - alk + oxide. The other choice would be alkan + oxide = alkanoxide. We decided that alkoxide is better. Back
Tutorial 8.4
Carbonium ion can give rearrangement products, and also elimination products. Reference Back
Tutorial 8.5
Alcohol is a weak acid and pyridine a base. They react to give RO‾[C5H5NH]+. The displacement of chloride from the benzenesulfonate chloride is easier with an alkoxide than an alcohol. Back
Tutorial 9.1
4-chloro-2-methoxyl-2-methylheptane Back
Reaction (1). Reaction (2) will also execute a E2 Reaction to give methanol and propylene. BackTutorial 9.2
The tert-butanol will yield a tertiary carbonium ion. Since the temperature used is relatively mild the carbonium ion will attack the oxygen of the ethanol displacing the proton to give the ether. The tert-carbonium ion is not likely to attack the oxygen of the tert-butanol because of steric hindrance. BackTutorial 9.3
Trans-2,3-dimethyloxirane. The SN2 attack by the alkoxide upon the carbon is from the rear with respect to the leavng chloride. BackTutorial 9.4
Tutorial 9.5
�
H2O
�
HBr
�
Back
Tutorial 10.1
4-amino-2-methylhexane, 1-(methylamino)-3-methylpentane. Back
Tutorial 10.2
CH2=CHCH2CH2CH2−N(CH3)2 ; N,N-dimethyl-pent-4-en-1-amine. Back
Tutorial 10.3
| NH3 + CH3CHO | � | CH3CH=NH |
H2 / Cat � |
CH3CH2NH2 |
| CH3CH2NH2 + CH3CHO | � | CH3CH=NCH2CH3 |
H2 / Cat � |
CH3CH2−NH−CH2CH3 Back |
Tutorial 11.1
Add in a little deuterium, an isotope of proton, into the mixture. Being an isotope it has the same chemistry as the proton. So we will end up with some ketone molecules with deuterium in them. The presence of such isotope can be easily detected. Alternatively use an enantiomer, like (R)-2-methyl-2-pentanone. Dissolve this in water and then recovered the compound. You will find that the optical rotation will have decreased owing to the breaking and rebonding of the proton at the α−carbon. A reaction that only convert an enantiomerically homogeneous compound into a racemate (that is a mixture with both enantiomers) is known as racemization. Back Tutorial 11.3
2-ethylhex-2-enal; 1-acetyl-2-methylcyclopentene; 2-(2-phenylethenyl)-cyclohexanone Back Tutorial 11.4
You have to use a 3-D model to see that one side of the carbonyl sp�−plane is hindered by a methyl substituent while the other side is by the cyclohexyl substituent, in the most stable conformation of the aldehyde. So the more frequent reaction will be from the lesser hindered methyl side. Back Tutorial 11.5
This is similar to tutorial 11.5, except the cyclic structure held the conformation about the carbonyl rigidly. Just like in a fight a friend held-back your oponent and you keep punching him. So the side hindered by only a proton sees 90% of all the action. Back Tutorial 11.6
Ph3P + CH3Br � [Ph3P−CH3]+Br- + BuLi � Ph3P=CH2 + BuH + LiBr Tutorial 11.7
Ph3P=CH2 + cyclohexanone � methylenecyclohexane + Ph3P=O
The reactions are conducted in ether. Back
Dry the leaves and ground it into fine powder. Stir it in an alkaline aqueous solution. Filter of the leave. Acidify the filtrate and extract it with ether. Remove the ether and you will get the acid. Back Tutorial 12.1
3-methylpentanoic acid. Back Tutorial 12.2
Both the carbon−oxygen bond length are equal (1.26 A˚), which is between the bond length of 1.20 A˚ for C=O and 1.34 A˚ for C−O. Back Tutorial 12.3
Highly electronegative atoms, like chlorine, tend to withdraw electron density from the carbon. This would favour the formation of an anion. Back Tutorial 12.4
