Aldehydes and ketones having
α-hydrogen atoms, undergoes aldol condensation, in present of base (NaOH).
The initial product formed during this reaction is
β-hydroxy alcohol, which then undergoes dehydration to form
α,β-unsaturated aldehyde or ketone.
In present case, 3,3-dimethyl-2-butanone has 2α-hyrogen atom, while methylcyclopentane-1-carbaldehyde has 1α-hydrogen atom. So the major product formed during cross aldol condensation reaction of these reactants is:
5-hydroxy-4,4-dimethly-1-(2-methylcyclopentyl)pent-1-en-3-one. The complete reaction product formed is shown below.
I don’t see any questions
Use the Ideal Gas Law to find the moles of gas first.
Be sure to convert T from Celsius to Kelvin by adding 273.
Also I prefer to deal with pressure in atm rather than mmHg, so divide the pressure by 760 to get it in atm.
PV = nRT —> n = PV/RT
P = 547 mmHg = 547/760 atm = 0.720 atm
V = 1.90 L
T = 33°C = 33 + 273 K = 306 K
R = 0.08206 L atm / mol K
n = (0.720 atm)(1.90 L) / (0.08206 L atm / mol K)(306 K) = 0.0545 mol of gas
Now divide grams by mol to get the molecular weight.
3.42 g / 0.0545 mol = 62.8 g/mol
Use equation
number of moles= sample mass/molar mass
Using the Henderson-Hasselbalch equation on the solution before HCl addition: pH = pKa + log([A-]/[HA]) 8.0 = 7.4 + log([A-]/[HA]); [A-]/[HA] = 4.0. (equation 1) Also, 0.1 L * 1.0 mol/L = 0.1 moles total of the compound. Therefore, [A-] + [HA] = 0.1 (equation 2) Solving the simultaneous equations 1 and 2 gives: A- = 0.08 moles AH = 0.02 moles Adding strong acid reduces A- and increases AH by the same amount. 0.03 L * 1 mol/L = 0.03 moles HCl will be added, soA- = 0.08 - 0.03 = 0.05 moles AH = 0.02 + 0.03 = 0.05 moles Therefore, after HCl addition, [A-]/[HA] = 0.05 / 0.05 = 1.0 Resubstituting into the Henderson-Hasselbalch equation: pH = 7.4 + log(1.0) = 7.4, the final pH.