Tutor profile: Nicky J.
Subject: Organic Chemistry
An ester has been synthesised. C2H5COOC3H7 is the semi-structural formula. 1a. Give the name of the ester, assuming the chains a linear. 1b. Name the reagents required to synthesise the ester. 1c. How could the reaction yield be improved. 1d. Describe the 1H NMR spectrum of the ester
1a. Naming an ester comes in 2 parts. The (C2H5CO) and (OC3H7). The (OC3H7) part comes first which will end in -yl, because there is 3 carbons it will be propyl. Then the (C2H5CO) part second which ends in -oate due to the carbonyl, and again there is 3 carbons, it will be propanoate. Therefore the ester will be propyl propanoate. 1b. To synthesise an ester, this requires a carboxylic acid and alcohol. The propyl of the ester will be the alcohol derivative and the propanoate will be the carboxylic acid derivative. Therefore propanol and propanoic acid will be the reagents required to synthesise the ester. 1c. The esterification reaction is an equilibrium. To increase efficiency of the reaction, a couple of things can be done. An acid catalyst can be added in order to increase the electrophilicity of the carbonyl, drawing the electron density from the carbon. Instead of reacting the carboxylic acid, the acid chloride can be synthesised by reacting with SOCl2. The acid chloride (and the acid anhydride) derivatives of the carboxylic acid more polarised, therefore are more reactive to an electrophilic attack which is the rate limiting step. 1d. First to consider is the amount of hydrogen environments. Each hydrogen is in different environment as of the asymmetry of the molecule. Therefore the NMR will have 5 peaks. The 2 methyl groups will have peaks ~1-2 ppm and an integration of 3. Due to the 2 hydrogens being on adjacent carbons these peaks will appear as triplets. The hydrogens on the carbon adjacent to the carbonyl will have a peak ~2-3 ppm and integration of 2. With the methyl group next adjacent the peak will appear as a quartet. COOCH2, These hydrogens will be shifted downfield due to the oxygen adjacent deshielding. This peak will be around 4 ppm and appear as a triplet due to the adjacent carbons 2 hydrogens. The remaining hydrogens environment will be more deshielded than the methyl's, being closer to the oxygen. Therefore, the peaks ppm will be higher than the methyl's at around 1.5-2 ppm. Also due to the hydrogen's having adjacent hydrogens (2 and 3) on either side, this will appear as a multiplet.
Subject: Basic Math
A designer shop has a 20% sale for shoes and 30% sale for trousers. The original price of the shoes are $59 and the original price of the trousers are $42. The price of the shoes and 2 pairs of trousers in a cheaper store combined are $105. Which shop is the cheaper option to get a pair of shoes and 2 pairs of trousers.
When working out percentages, this can be done a few ways. If you can use a calculator, you can work out the price of the designer shoes by. 1.0 (100%) - 0.2 (20% off) = 0.8 (80% of original value) 0.8 * $59 = $47.20 If a calculator is not allowed, to get to 80% of the original value it is easiest to divide the original value by 10 to get to 10% then times by 8 to get to 80% of the original value. $59 / 10 = $5.90 * 8 = $47.20 This can be repeated for the 30% off for the trousers in either way. But account this is for 70% original value. 0.7 * $42 = $29.40 $42 / 10 = $4.20 * 7 = $29.40 Then to get the total value of a pair of shoes and 2 pairs of trousers. $47.20 for one pair of shoes need to added to 2 * $29.40 (2*29.40 = $58.80). $47.20 + $58.80 = $106.00 $106.00 is the total for the designer store, for the pair of shoes and 2 pairs of trousers. Therefore, it is cheaper to get the items in the cheaper store than in the designer store with discount by $1.00 If you require further assistance with multiplication, addition, division or subtraction just ask.
1a. In terms of orbitals, compare and explain the differences bonding and bond lengths in the nitrogen, oxygen and fluorine diatomic molecules.
When these kind of questions are put in front of you i.e. to compare and contrast different elements and how they bond, it is important to look at periodicity. Nitrogen, oxygen and fluorine are all in the 2nd row and have the same 2p orbital. Then you see the difference between the elements is the number of electrons, increasing in number down the period. So nitrogen [He]2s2 2p3, oxygen [He]2s2 2p4, fluorine [He]2s2 2p5. The increased number of electrons contract the orbital due to the increased negative charge of the electrons being attracted to the positive nucleus further. Therefore, typical sigma bonding strength between the diatomic molecules increases down the period. Therefore F-F > O-O > N-N in terms of bond strength. However, O-O and N-N diatomic molecules do not just form sigma bonds but also pi bonds. Down to the octet rule fluorine has 7 electrons in its outer shell and the two fluorine molecules are happy to donate an electron each to form a covalent bond and form a sp3 hybridised orbital. This is not the case for nitrogen and oxygen which require more than one electron donation to satisfy the octet rule. Oxygen needing 2 and nitrogen needing 3. To then satisfy this oxygen forms a double bond (each oxygen donating 2 electrons in a sp3 and sp2 hybridised orbitals) and nitrogen forms a triple bond (each nitrogen donating 3 electrons in a sp3, sp2, sp hybridised orbitals). Due to the increasing strength of the diatomic molecules bonding with double and triple bonds the N2 has the strongest bond followed by O2 and then F2. To satisfy the question requirements the stronger the bond, the shorter the bond length. So the shortest bond length is N2 then O2 then F2.
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