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.
By definition, one mole (one gram molecular weight) of any substance, contains Avogadro’s number of particles; atoms if you are discussing an element, or molecules if a compound. Avogadro’s number has been determined by several methods, all of the accepted values lie within a range of +-1% about the value of 6.022045 x 10^23/gm. That is a large number, in this case approximately; 602,204,500,000,000,000,000,000 molecules of glucose.
From the web :v
Answer:
0.46 grams (C₆H₅)₂CO
Explanation:
To find the mass of benzophenone ((C₆H₅)₂CO), you need to (1) convert mmoles to moles and then (2) convert moles to grams (via molar mass). It is important to arrange the conversions/ratios in a way that allows for the cancellation of units. The final answer should have 2 sig figs to match the sig figs of the given value (2.5 mmoles).
Molar Mass ((C₆H₅)₂CO): 13(12.011 g/mol) + 10(1.008 g/mol) + 15.998 g/mol
Molar Mass ((C₆H₅)₂CO): 182.221 g/mol
2.5 mmoles (C₆H₅)₂CO 1 mole 182.221 g
----------------------------------- x ------------------------ x ------------------- =
1,000 mmoles 1 mole
= 0.46 grams (C₆H₅)₂CO
Graduated cylinder
Stopwatch
Thermometer