Tutor profile: Baylee T.
Subject: Organic Chemistry
An optically active, halogen-containing alkane was treated with concentrated methanol (MeOH). The resulting product was optically inactive but gave sharp IR peaks suggesting the presence of a ether bond. What classification of reaction occurred?
The classification of reaction that occurred was a substitution uni-molecular (Sn1) reaction. No significant information is given about the substitution of the substrate, only that it is halogenated; while the level of substitution would be nice to know, what is really important here is that the reactant contained a good leaving group (the halide). Treating our halogenated alkane with a weak nucleophile caused the formation of a species that was BOTH optically inactive and a species that contained an ether group. The optical inactivity would caused by the formation of a racemic mixture often experienced in an Sn1 reaction while the presence of the ether group is a result of the substitution of the methanol group where the leaving group was.
To determine the concentration of a hydrochloric acid (HCl) solution, a 50mL aliquot of HCl was titrated against a primary standard of 5M NaOH. After adding a color-based pH indicator, HCl was added dropwise into 100mL of NaOH. A color change was observed after adding 45mL of HCl. What was the concetration of the HCl stock solution?
The color change reaction observed after the addition of 45mL of HCl is called the "equivalence point" (Veq). This point is defined by when the concentration of hydrogen and hydroxyl ions in solution are equal. Due to the fact we are dealing with both a strong base and a strong acid, this point can be used to directly calculate the concentration of the HCl solution. 1.) Use the molarity of the NaOH to determine the moles of NaOH in the initial solution using the molarity formula. - 5M= Xmoles NaOH/ 0.100L solution gives us 0.5moles NaOH in 100mL. 2.) Understanding the NaOH is a strong base allows us to assume that the number of moles of NaOH = the number of moles of OH- in solution. - 0.5moles NaOH= 0.5moles OH- in solution. 3.) The color change indicated we were at Veq so OH-=H+. - 0.5moles OH-= 0.5moles H+. 4.) Since HCl is also a strong acid, we can assume that the moles of H+ in solution = the moles of HCl in solution. - 0.5moles H+ = 0.5moles HCl 5.) Knowing that we have 0.5moles of HCl in solution allows us to determine the concentration of the HCl by factoring in the added volume of 45mL. This can be done using the molarity formula again. - XM= 0.5mL / 0.045L gives us 11.1M HCl.
messenger ribonucleic acid (mRNA) undergoes multiple processing steps before the mRNA transcript leaves the nucleus. An mRNA transcript was sequenced and was found to contain several inter-genetic regions that did not code for protein. What step in mRNA processing step did not occur on this transcript?
The mRNA processing step known as "RNA splicing" did not occur. During this step, multiple mRNA transcripts have their inter-genetic regions (introns) cleaved and removed from the transcript. The fragments that are now missing introns are then ligated together to form a string of gene containing mRNA transcripts.
needs and Baylee will reply soon.